From the given information, we can derive that \begin{align*} f(x) &= f(2158 - x) = f(3214 - (2158 - x)) = f(1056 + x) \\ &= f(2158 - (1056 + x)) = f(1102 - x) \\ &= f(1102 - (1056 + x)) = f(46 - x) \\ &= f(398 - (46 - x)) = f(352 + x).\end{align*}It follows that $ f(x) $ is periodic, whose period divides 352. This means that every value in the list $ f(0), $ $ f(1), $ $ \dots, $ $ f(999) $ must appear among the values \[f(0), f(1), f(2), \dots, f(351).\]The identity $ f(x) = f(398 - x) $ implies that every value in the list $ f(200), $ $ f(201), $ $ \dots, $ $ f(351) $ must appear among the values \[f(0), f(1), \dots, f(199),\]and the identity $ f(x) = f(46 - x) $ implies that every value in the list $ f(0), $ $ f(1), $ $ \dots, $ $ f(22) $ must appear among the values \[f(23), f(24), \dots, f(199).\]This implies that $ f(23), $ $ f(24), $ $ \dots, $ $ f(199) $ capture all the possible values of $ f(n), $ where $ n $ is a positive integer. Now, let $ f(x) = \cos \left( \frac{360}{352} (x - 23) \right), $ where the cosine is evaluated in terms of degrees. Then \[1 = f(23) > f(24) > f(25) > \dots > f(199) = -1,\]and we can verify that $ f(x) = f(398 - x), $ $ f(x) = f(2158 - x), $ and $ f(x) = f(3214 - x) $. Thus, the list $ f(0), $ $ f(1), $ $ \dots, $ $ f(999) $ can have at most $ 199 - 23 + 1 = \boxed{177} $ different values.